Respuesta :
Answer : The age of sample is [tex]9.33\times 10^3\text{ years}[/tex]
Explanation :
Half-life = 5720 years
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{5720\text{ years}}[/tex]
[tex]k=1.21\times 10^{-4}\text{ years}^{-1}[/tex]
Now we have to calculate the age of sample.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]1.21\times 10^{-4}\text{ years}^{-1}[/tex]
t = age of the sample = ?
a = let initial amount of the reactant = 13.6 counts per min
a - x = amount left after decay process = 4.4 counts per min
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{13.6}{4.4}[/tex]
[tex]t=9327.8\text{ years}=9.33\times 10^3\text{ years}[/tex]
Therefore, the age of sample is [tex]9.33\times 10^3\text{ years}[/tex]
The age of the wood sample that gives a rate for carbon-14 decay of 4.4 counts per minute is 9.33 x 10³ years.
What is the rate constant for the process?
We know the formula for the rate constant, k is given as,
[tex]k = \dfrac{0.693}{t_{(1/2)}}[/tex]
Half-life, [tex]t_{(1/2)} = 5720\ year[/tex]
[tex]k = \dfrac{0.693}{5720}\\\\k = 1.2115\times 10^{-4}\rm\ year^{-1}[/tex]
What is the age of the sample?
We know that the expression for rate law for first-order kinetics is given by,
[tex]t = \dfrac{2.303}{k}\rm log\dfrac{a}{a-x}[/tex]
Half-life, [tex]t_{(1/2)} = 5720\ year[/tex]
Rate constant, k = 1.21 x 10⁻⁴ year⁻¹
The initial amount of sample, a = 13.6 counts per min
Amount after decay process, (a-x) = 4.4 counts per min
[tex]t = \dfrac{2.303}{1.2115\times 10^{-4}\rm\ year^{-1}}\rm log\dfrac{13.6}{4.4}\\\\t = 9.33 \times 10^3\ years[/tex]
Hence, the age of the wood sample that gives a rate for carbon-14 decay of 4.4 counts per minute is 9.33 x 10³ years.
Learn more about Carbon-14 decay:
https://brainly.com/question/1500129
